1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Finding the graph formula with know points and other equation.

  1. Mar 11, 2014 #1
    I have an equation for determining the acceleration of an object being propelled by a constant power source, that is affected by air resistance:

    a = [itex]\frac{P}{mv}[/itex]-[itex]\frac{CDpAv2}{2m}[/itex]

    Since F = [itex]\frac{P}{v}[/itex]

    I am trying to graph this as a velocity-time graph, however, I don't know how to do it. There is no time variable that I can replace with x, and the y-value (velocity) is mixed into the equation already.

    I remember the equation given to me for a similar sort of thing, without air resitance, but instead just a constant friction force, that was something like this:

    y = k(1-e-ax)

    Where k is a constant, which is the maximum speed, and a is another constant which represents the force of air resistance.

    The maximum speed in this case is [itex]\sqrt[3]{\frac{2P}{CDpA}}[/itex], so the equation would be something like:

    y = [itex]\sqrt[3]{\frac{2P}{CDpA}}[/itex](1-e-ax)

    But that's as far as I've gotten. By the use of iteration, I have determined the velocity at several different times, here's a few, just in case it helps:

    (0,0) (25,83.4762) (50,118.1195) (75,126.1601) (100,127.7624) (125,128.0709)

    This is for variables of values: p = 0.001, A = 1900.933, m = 1900.933, P = 1*106
    and CD = 0.5

    Any help would be greatly appreciated, and if you need any more information, just let me know.

    EDIT: the equations don't seem to be formatting correctly, so I'll redo them down here:

    a = P/v - (Cd*p*A*v^2) /2
    F = P/v
    max speed = ( (2*P) / (Cd*p*A) )^1/3
    y = ( (2*P) / (Cd*p*A) )^1/3 * (1 - e^-ax)
    Last edited: Mar 11, 2014
  2. jcsd
  3. Mar 11, 2014 #2
    Given ##a=\frac{dv}{dt}##, what you have is a separable ordinary differential equation [tex]\frac{dv}{dt}=\frac{\alpha}{v}-\beta v^2[/tex] where ##\alpha=\frac{P}{m}## and ##\beta=\frac{C_DpA}{2m}## have been substituted to make the equation (slightly) more manageable. Hopefully that's enough to get you going.

    I'm not sure where you got your force equation. Are you not using ##F=ma##? Furthermore, I don't know that it's even useful here.

    I reckon it's doubtful that the equation that works in a constant friction situation applies here. So you're likely barking up the wrong tree there.
  4. Mar 12, 2014 #3
    I have never done differential equations before, and I tried to just find the anti-derivative, but my teacher said that I had done it incorrectly, as I differentiated dv instead of dt, or the other way around. Anyway, here is what I got:

    v = (Pt/m)*ln(|v|)-(CDpAv3t)/(6m)+c

    Unfortunately from there, if it is correct, I don't know where to go - how to get v by itself without v on the other side.

    What do I do from here, or from the start if that isn't correct?

    Thanks for your response.
    Last edited: Mar 12, 2014
  5. Mar 12, 2014 #4

    Stephen Tashi

    User Avatar
    Science Advisor

    I think this is due to your using the SUB and SUP tags instead of the usual LaTex notation.

    [itex] a = \frac{P}{v} - \frac{ (C_d p A v^2) }{2} [/itex]
  6. Mar 12, 2014 #5
    Oh ok, I'll give it a go:

    a = ([itex]\frac{Pt}{m}[/itex])ln(|v|)-[itex]\frac{C_{D}pAv^{3}t}{6m}[/itex]

    Seems to work, thanks.
  7. Mar 16, 2014 #6
    Does anyone have any advice on what I should do to get time into the equation, without velocity on both sides? I don't even know what to type into google to find information on it. Any information would be much appreciated.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook